Codeforces Round #218 (Div. 2) 题解

  1. K-Periodic Array

题意:有n个数,有一个k,问最少修改几个数,使的a[i%k+1]都相等

思路:对每个位置都统计一下每种数字的出现次数,取最大值

代码:

#include <bits/stdc++.h>
using namespace std;

int n,k;
int a[105];

int main()
{
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) {
        scanf("%d",&a[i]);
    }
    int ans=0;
    for(int i = 1; i <= k; i++) {
        int cnt = 0,maxe = 0;
        map<int,int> mp;
        for(int j = i; j <= n; j += k) {
            mp[a[j]]++;cnt++;
            if(mp[a[j]]>maxe)maxe=mp[a[j]];
        }
        ans+=cnt-maxe;
    }
    cout<<ans<<endl;
    return 0;
}
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B. Fox Dividing Cheese

    题意:给出ab两个数字,每次可以对其中的一个数字*1/2,或者*1/3,或者*1/5,问最少几次可以使ab相等

    思路:对ab进行2 3 5分解,看最后分解剩下的是否相等,如果不相等就打印-1,相等减去找到最小值

    代码:

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    int a2=0,a3=0,a5=0,b2=0,b3=0,b5=0;
    while(a%2==0)a/=2,a2++;
    while(a%3==0)a/=3,a3++;
    while(a%5==0)a/=5,a5++;
    while(b%2==0)b/=2,b2++;
    while(b%3==0)b/=3,b3++;
    while(b%5==0)b/=5,b5++;
    if(a!=b){
        printf("-1\n");
    }
    else{
        printf("%d\n",abs(a2-b2)+abs(a3-b3)+abs(a5-b5));
    }
    return 0;
}
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C. Hamburgers

    题意:汉堡有BSC三种材料构成,先输入汉堡的配方,然后给出每种原料现在有多少,以及每种原料超市中的单价,以及你现在有多少钱,问最多可以做多少个汉堡

    思路:对答案进行二分,然后判断做这么多汉堡需要多少钱

    代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

char str[105];
LL cntb,cnts,cntc;
LL numb,nums,numc;
LL valb,vals,valc;
LL a;
bool check(LL x)
{
    LL res = 0;
    LL as = x * cntb - numb;
    if(as > 0) res += as * valb;
    as = x * cnts - nums;
    if(as > 0) res += as * vals;
    as = x * cntc - numc;
    if(as > 0) res += as * valc;
    if(res <= a)return true;
    return false;
}
int main()
{
    scanf("%s",str+1);
    cin>>numb>>nums>>numc;
    cin>>valb>>vals>>valc;
    cin>>a;
    int len = strlen(str+1);
    cntb = 0; cntb = 0; cntc = 0;
    for(int i = 1; i <= len; i++) {
        if(str[i] == 'B')cntb++;
        if(str[i] == 'S')cnts++;
        if(str[i] == 'C')cntc++;
    }
    LL l = 0,r = 1e15;
    LL ans = 0;
    while(l <= r){
        LL mid =(l + r) >> 1;
        if(check(mid)){
            l = mid + 1;
            ans = mid;
        }
        else r = mid - 1;
    }
    cout<< ans <<endl;
    return 0;
}
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D. Vessels

    题意:有一个n(2e5)层的容器,每一层的容量为ai,每一层满了以后就会自动流入到下一层,有两种操作,一种是向第i层加入xi的水,第二种操作是查询第x层现在有多少水,

思路:用一个类似于跳表的东西维护每一层如果满了下一层流向哪里

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn=2e5+7;
int a[maxn],now[maxn];
int fa[maxn];
int n;

int update(int id,int val)
{
    if(now[id] + val <= a[id]) {
        now[id] += val;
        return id;
    }
    val = now[id] + val - a[id];
    now[id] = a[id];
    fa[id] = update(fa[id],val);
    return fa[id];
}

int main()
{
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) {
        scanf("%d",&a[i]);
        fa[i] = i+1;
    }
    a[n+1]=1e15;
    int q;
    scanf("%d",&q);
    while(q--){
        int op;
        scanf("%d",&op);
        if(op==1) {
            int as, qw;
            scanf("%d%d",&as,&qw);
            update(as, qw);
        }
        else {
            int as;
            scanf("%d",&as);
            printf("%d\n",now[as]);
        }
    }
    return 0;
}
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E. Subway Innovation

    题意:在数轴上有n(3e5)个点,从中选出k个点,使的这k个点的价值最小,k个点的价值为这k个点中任意两个点的距离之和

    思路:对数轴上的点排序,因为让最后答案最小,所以最后的点集肯定是数轴上连续的k个点,假设选了[l,r]的k个点,与[l+1,r+1]的关系是

 

所以排序以后,直接On维护求出最小值即可

  代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn=3e5+7;
struct node
{
    int id;
    LL val;
    bool operator <(const node &b)const {
        return val < b.val;
    }
};
node a[maxn];
LL pre[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) {
        LL x;
        scanf("%lld",&x);
        node as;
        as.id = i; as.val = x;
        a[i] = as;
    }
    int K;
    scanf("%d",&K);
    sort(a + 1, a + 1 + n);
    for(int i = 1; i <= n; i++) {
        pre[i] = pre[i - 1] + a[i].val;
    }
    LL res = 0;
    for(int i = 1; i <= K; i++) {
        res += (i - 1) * a[i].val - pre[i - 1];
    }
    LL maxe = res;
    int k=1;
    for(int i = K + 1; i <= n; i++) {
        res += (K - 1) * (a[i].val + a[i - K].val) - 2 * (pre[i - 1] - pre[i - K]);
        if(res < maxe) {
            maxe = res; k = i - K + 1;
        }
    }
    for(int i = k; i < K + k; i++) {
        printf("%d ",a[i].id);
    }
    puts("");
    return 0;
}
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posted @ 2019-02-14 15:45  啦啦啦天啦噜  阅读(206)  评论(0编辑  收藏  举报